# Parametric Integrals

A LevelAQAEdexcelOCR

## Parametric Integrals

When dealing with parametric equations, integrals become more complicated. We cannot just do $\int y \, dx$ when we don’t have $y$ written in terms of $x$. Instead, we must use the chain rule to get an integral in terms of the parameter. Then, if it is a definite integral, we must convert the limits to fit the new integration.

Make sure you are happy with the following topics before continuing.

A LevelAQAEdexcelOCR

## The Chain Rule

Recall: The Chain Rule.

$\dfrac{dy}{dx}=\dfrac{dy}{dz}\dfrac{dz}{dx}$

If we have parametric equations and $y$ isn’t written in terms of $x$, but instead it is written in terms of $t$ say, then we can use the chain rule to show that $dx=\dfrac{dx}{dt} \, dt$ for a parameter $t$, and since we have $x$ in terms of $t$ we can get $\dfrac{dx}{dt}$ in terms of $t$, and we already have $y$ in terms of $t$, so our integral can be written as:

${\LARGE \int} y \, dx={\LARGE \int} y \, \dfrac{dx}{dt} \, dt$

A LevelAQAEdexcelOCR

## Limit Conversion

If we have a definite integral $\int^{b}_{a}y \, dx$, then we cannot just take our limits $a$ and $b$ and put them on our new integral in terms of $t$, because they are limits with respect to $x$.

Instead, we need to convert them.

This means that the lower limit on the integral in terms of $t$ is the $t$ value that gives $x=a$, and the upper limit on the integral in terms of $t$ is the $t$ value that gives $x=b$.

With these converted limits we can find the value of the definite integral.

A LevelAQAEdexcelOCR

@mmerevise

A LevelAQAEdexcelOCR

## Example 1: Using the Chain Rule

A parametric equation is $x=3t+4$ and $y=t^{2}$. Find $\int y \, dx$ in terms of $t$.

[2 marks]

$\int y \, dx={\LARGE \int} y\dfrac{dx}{dt} \, dt$

$x=3t+4$

$\dfrac{dx}{dt}=3$

$y=t^{2}$

$\int y \, dx=\int 3t^{2} \, dt$

$\int y \, dx=t^{3}+c$

A LevelAQAEdexcelOCR

## Example 2: Definite Integrals

A parametric curve is defined by $y=t^{3}+3t$, $x=t^{2}+4t+4$, for $t>-3$. Find $\int^{4}_{0}y \, dx$.

[3 marks]

First convert limits.

$x=t^{2}+4t+4$

First limit: $x=0$

$t^{2}+4t+4=0$

$(t+2)^{2}=0$

$t=-2$

Second limit: $x=4$

$t^{2}+4t+4=4$

$t^{2}+4t=0$

$t(t+4)=0$

$t=0$ or $t=-4$

$t=-4$ not in range

$t=0$

$\int^{4}_{0} y \, dx={\LARGE\int}^{0}_{-2} \, y\dfrac{dx}{dt} \, dt$

$x=t^{2}+4t+4$

$\dfrac{dx}{dt}=2t+4$

$y=t^{3}+3t$

\begin{aligned}\int^{4}_{0} y \, dx&=\int^{0}_{-2}(t^{3}+3t)(2t+4) \, dt\\[1.2em]&=\int^{0}_{-2}\left( 2t^{4}+4t^{3}+6t^{2}+12t\right) dt\\[1.2em]&=\left[\dfrac{2}{5}t^{5}+t^{4}+2t^{3}+6t^{2}\right]^{0}_{-2}\\[1.2em]&=\left( \dfrac{2}{5}\times0^{5}\right) +0^{4}+\left( 2\times0^{3}\right) +\left( 6\times0^{2}\right)-\left( \dfrac{2}{5}\times(-2)^{5}\right) -(-2)^{4} - 2(- 2)^{3} - 6( -2)^{2} \\[1.2em]&=\left( \dfrac{2}{5}\times32\right) -16+\left( 2\times8\right) -\left( 6\times4\right) \\[1.2em]&=\dfrac{64}{5}-16+16-24\\[1.2em]&=-\dfrac{56}{5}\end{aligned}

A LevelAQAEdexcelOCR

## Parametric Integrals Example Questions

$\int y \, dx={\LARGE \int} y\dfrac{dx}{dt}dt$

$x=t^{2}+2$

$\dfrac{dx}{dt}=2t$

$y=t^{3}+4t^{2}+4t+3$

\begin{aligned}\int y \, dx&=\int\left( 2t(t^{3}+4t^{2}+4t+3)\right) \, dt \\[1.2em]&=\int\left( 2t^{4}+8t^{3}+8t^{2}+6t\right) \, dt \end{aligned}

$\int y \, dx={\LARGE \int} y\dfrac{dx}{dt}dt$

$x=4t^{\frac{5}{9}}$

$\dfrac{dx}{dt}=\dfrac{20}{9}t^{-\frac{4}{9}}$

$y=t^{-\frac{1}{2}}$

\begin{aligned}\int ydx&=\int \left( t^{-\frac{1}{2}}\times\dfrac{20}{9}t^{-\frac{4}{9}}\right) dt\\[1.2em]&=\int \dfrac{20}{9}t^{-\frac{17}{18}}dt\\[1.2em]&=\left( \dfrac{20}{9}\div\dfrac{1}{18}\right) t^{\frac{1}{18}}+c\\[1.2em]&=40t^{\frac{1}{18}}+c\end{aligned}

First find the new limits.

Upper limit $x=6$

$2t+4=6$

$2t=2$

$t=1$

Lower limit $x=0$

$2t+4=0$

$2t=-4$

$t=-2$

Thus:

$\int^{6}_{0}y \, dx={\LARGE \int}^{1}_{-2}y\dfrac{dx}{dt} \, dt$

$x=2t+4$

$\dfrac{dx}{dt}=2$

$y=t+6$

\begin{aligned}\int^{6}_{0}y \, dx&=\int^{1}_{-2}2(t+6) \, dt\\[1.2em]&=\int^{1}_{-2}\left( 2t+12\right) \, dt\\[1.2em]&=[t^{2}+12t]^{1}_{-2}\\[1.2em]&=1^{2}+\left( 12\times1\right) -(-2)^{2}-\left( 12\times(-2)\right) \\[1.2em]&=1+12-4+24\\[1.2em]&=33\end{aligned}

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